= Solution
Take the sky frame and disk frame to be right-handed <orthonormal bases>, with positive angular displacement measured from $\widehat X$ toward $\widehat Y$. In particular, positive $Z$ is toward the observer, so material crossing the sky plane with positive $\dot Z$ is at the <ascending node>. Write $c_d=\cos\Omega_d$, $s_d=\sin\Omega_d$, $C_d=\cos I_d$, and $S_d=\sin I_d$. The disk frame's first <unit vector> lies along its <ascending node>, and its second has positive $Z$ component. The <orbital-frame rotation from inclination and node> therefore gives
$$
\boxed{T=R_Z(\Omega_d)R_X(I_d)=
\begin{pmatrix}
c_d&-s_dC_d&s_dS_d\\
s_d&c_dC_d&-c_dS_d\\
0&S_d&C_d
\end{pmatrix}.}
$$
Each column is a disk-frame <unit vector> in sky coordinates, so $\mathbf X=T\mathbf x$ and the inverse is $\mathbf x=T^{\mathsf T}\mathbf X$. At the disk's <ascending node>, the $Z$ component of its tangential <velocity> is positive, confirming the sign of the tilt.
Let $T_p=R_Z(\Omega_p)R_X(I_p)$ define a provisional planet frame whose first axis is the planet's sky-plane <ascending node>. Its normal is $\widehat z_p=(\sin\Omega_p\sin I_p,-\cos\Omega_p\sin I_p,\cos I_p)$. Expressing that normal in the disk <orthonormal basis>, with $\Delta\Omega=\Omega_p-\Omega_d$, gives
$$
T^{\mathsf T}\widehat z_p=
\begin{pmatrix}
\sin I_p\sin\Delta\Omega\\
-\cos I_d\sin I_p\cos\Delta\Omega+\sin I_d\cos I_p\\
\cos I_d\cos I_p+\sin I_d\sin I_p\cos\Delta\Omega
\end{pmatrix}.
$$
The last component is the <dot product> of the two orbital normals, hence the cosine of the <mutual inclination>. The mutual <ascending node> points along $\widehat z_d\times\widehat z_p$: motion there has positive disk-normal component. Set
$$
A=\cos I_d\sin I_p\cos\Delta\Omega-\sin I_d\cos I_p,
\qquad B=\sin I_p\sin\Delta\Omega.
$$
Then $\widehat x\prime=(A\widehat x+B\widehat y)/\sqrt{A^2+B^2}$, and the quadrant-correct <longitude of ascending node> is $\Omega_m=\operatorname{atan2}(B,A)$. Thus
$$
\boxed{\cos I_m=\cos I_d\cos I_p+\sin I_d\sin I_p\cos\Delta\Omega,
\qquad \tan\Omega_m=\frac{\sin I_p\sin\Delta\Omega}{\cos I_d\sin I_p\cos\Delta\Omega-\sin I_d\cos I_p}.}
$$
The tangent alone cannot fix the quadrant, and $\Omega_m$ is undefined for exactly coincident <orbital planes>.
For the two rotation routes, define $\psi$ as the oriented angle, within the planet's <orbital plane>, from its sky-plane <ascending node> to its disk-plane <ascending node>. If $\widehat x_p,\widehat y_p$ are the first two columns of $T_p$, take $\psi=\operatorname{atan2}(\widehat x\prime\cdot\widehat y_p,\widehat x\prime\cdot\widehat x_p)$. Both routes describe the same planet <orthonormal basis>:
$$
T R_z(\Omega_m)R_x(I_m)=T_pR_z(\psi).
$$
The left route goes through the disk frame; the right goes through the sky-node planet frame. Comparing their third columns gives the two boxed relations, while comparison of first columns fixes the extra in-plane angle.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-63-orbital-frames.png]
{title=Sky-node directions and the mutual ascending node of the disk and planet}
{height=440}
Put $I_d=\pi/2+\delta_d$ and $I_p=\pi/2+\delta_p$. A second-order <Taylor expansion> gives $\cos I_m=1-[(\delta_p-\delta_d)^2+\Delta\Omega^2]/2+O(\delta^4)$, and therefore the <mutual inclination of nearly edge-on orbits> satisfies
$$
\boxed{I_m^2=(I_p-I_d)^2+(\Omega_p-\Omega_d)^2+O(\delta^4).}
$$
All expansion angles are in radians. For plotting in degrees this is $I_m\simeq\sqrt{(I_p-87^\circ)^2+(4^\circ)^2}$. The given planet has exact $I_m\simeq4.119^\circ$ and $\Omega_m\simeq76.044^\circ$. The exact minimum occurs at $I_p=\operatorname{atan2}(\sin I_d\cos\Delta\Omega,\cos I_d)\simeq86.993^\circ$, with $I_{m,\min}\simeq3.995^\circ$. At $I_p=85^\circ$ and $95^\circ$, the second-order values are $\sqrt{20}^\circ\simeq4.472^\circ$ and $\sqrt{80}^\circ\simeq8.944^\circ$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-63-mutual-inclination.png]
{title=Mutual inclination versus the planet's observed inclination}
{height=340}
For the disk evolution, neglect disk self-gravity, collisions that align the planes, and back-reaction on the planet. In the initial disk plane, use the <complex inclination> $w=I e^{i\Omega}$ for a ring and $w_p=I_m e^{i\Omega_m}$ for the planet. Linear <Laplace-Lagrange secular theory> has <forced inclination> $w_p$ and solution $w(a,t)=w_p[1-e^{-i\nu(a)t}]$ for initially flat rings. For an inner circular perturber, the <Laplace coefficient> formula is $\nu=n(M_p/M_\star)\alpha b_{3/2}^{(1)}(\alpha)/4$, with $\alpha=a_p/a$. Its <quadrupole approximation> uses $b_{3/2}^{(1)}\simeq3\alpha$, yielding
$$
\nu(a)\simeq\frac34n\frac{M_p}{M_\star}\left(\frac{a_p}{a}\right)^2\propto a^{-7/2}.
$$
In the $(I\cos\Omega,I\sin\Omega)$ plot referenced to the initial disk, every ring runs clockwise on the same circle centered at $w_p$, starting at the origin. Inner rings advance more quickly. If the axes are instead referenced to the planet's plane, the relative tilt is $w-w_p=-w_pe^{-i\nu t}$: these circles are centered at the origin, and the <mutual inclination> is constant. Both conventions describe the same <nodal precession>; they should not be mixed.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-63-disk-precession.png]
{title=Differential nodal precession in initial-disk and planet-plane inclination coordinates}
{height=360}
At an intermediate epoch, rings with $\nu t\gtrsim1$ have tilted substantially while outer rings with $\nu t\ll1$ remain near the original disk plane. This radial variation is a <planet-induced debris-disk warp>. The characteristic affected radius grows like $t^{2/7}$. Conservative <differential nodal precession> preserves each ring's tilt relative to the planet, so it does not by itself align every orbit; an unresolved inner region can acquire a mean plane near the planetary plane through <phase mixing>. The secular interpretation is also described in https://people.ast.cam.ac.uk/~wyatt/lecture2_planetarysystemdynamics.pdf[Wyatt's planetary dynamics lectures].
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